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Proof step 2 from proof two-03 (previous step)

2 * m ≤ 2 ⊦ m = 1, m = 0

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%one: ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult